Primitive of x by Arctangent of x
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Theorem
- $\ds \int x \arctan x \rd x = \frac {x^2 + 1} 2 \arctan x - \frac x 2 + C$
Proof 1
From Primitive of $x \arctan \dfrac x a$:
- $\ds \int x \arctan \frac x a \rd x = \frac {x^2 + a^2} 2 \arctan \frac x a - \frac {a x} 2 + C$
The result follows on setting $a = 1$.
$\blacksquare$
Proof 2
With a view to expressing the primitive in the form:
- $\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
\(\ds u\) | \(=\) | \(\ds \arctan x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac {\d u} {\d x}\) | \(=\) | \(\ds \frac 1 {x^2 + 1}\) | Derivative of $\arctan x$ |
and let:
\(\ds \frac {\d v} {\d x}\) | \(=\) | \(\ds x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds v\) | \(=\) | \(\ds \frac {x^2} 2\) | Primitive of Power |
Then:
\(\ds \int x \arctan x \rd x\) | \(=\) | \(\ds \frac {x^2} 2 \arctan x - \int \frac {x^2} 2 \paren {\frac 1 {x^2 + 1} } \rd x + C\) | Integration by Parts | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 \arctan x - \frac 1 2 \int \frac {x^2 \rd x} {x^2 + 1} + C\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2} 2 \arctan x - \frac 1 2 \paren {x - 1 \arctan x} + C\) | Primitive of $\dfrac {x^2} {x^2 + a^2}$ with $a = 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^2 + 1} 2 \arctan x - \frac x 2 + C\) | simplifying |
$\blacksquare$
Sources
- 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text {II}$. Calculus: Exercises $\text {XIV}$: $24$.